Question 1. Graphical Integration Challenge: $u$-Substitution and Integration by Parts
Let $F(x)=\int_{-1}^{x} f(t) \mathrm{d} t$ and $G(x)=\int_{0}^{x} g(t) \mathrm{d} t$, where the graphs of $f(x)$ and $g(x)$ on $[-2,4]$ are below.
(a) Evaluate $\int_{0}^{2}[F(x) f(x)-G(x) g(x)] \mathrm{d} x$.
HINT: Consider a $u$-substitution (possibly multiple substitutions). You may need to invoke Part I of the Fundamental Theorem of Calculus.
(b) Evaluate $\int_{0}^{1} f^{\prime}(G(x)) G(x) g(x)$.
HINT: Consider a $u$-substitution. You may also need to integrate by parts.

Step 1 ：Given that $F(x)=\int_{-1}^{x} f(t) \mathrm{d} t$ and $G(x)=\int_{0}^{x} g(t) \mathrm{d} t$, we are asked to evaluate the following integrals:

Step 2 ：(a) $\int_{0}^{2}[F(x) f(x)-G(x) g(x)] \mathrm{d} x$

Step 3 ：Using the Fundamental Theorem of Calculus, we can simplify this integral to $F(2)f(2) - F(0)f(0) - G(2)g(2) + G(0)g(0)$

Step 4 ：\(\boxed{F(2)f(2) - F(0)f(0) - G(2)g(2) + G(0)g(0)}\)

Step 5 ：(b) $\int_{0}^{1} f^\prime(G(x)) G(x) g(x) dx$

Step 6 ：Using integration by parts and $u$-substitution, we can simplify this integral to $[f(G(x))G(x)]_{0}^{1} - \int_{0}^{1} f(G(x)) g(x) dx$

Step 7 ：\(\boxed{[f(G(x))G(x)]_{0}^{1} - \int_{0}^{1} f(G(x)) g(x) dx}\)